Curve sketching practice question 23
Let . Find the -coordinate of its local minimum.
is a quotient, so which rule gives ? Once you have the critical points, classify them by the sign of , not by which one is smaller.
Quotient rule: . The critical points are the roots of the numerator.
There are two critical points, one on each side of the vertical asymptote at . Decide which is the minimum from the sign of , or from the shape: the branch to the right of the asymptote comes down from .
An exact expression, or a decimal to two decimal places.
Theory UBC Math 100 §7.2CLP-1 §3.6Key results
Worked solution
Strategy
Differentiate, solve the quadratic in the numerator, and classify the two critical points. The vertical asymptote separates a local maximum on the left from a local minimum on the right.
Working
Quotient rule.
Quadratic formula.
Falling then rising to the right of the asymptote: a local minimum at 1 + √2 ≈ 2.414 (value 2 + 2√2). Left of the asymptote, f′ > 0 for x < 1 − √2 and f′ < 0 between 1 − √2 and 1, so 1 − √2 is a local maximum.
Answer
The answer is 1 plus root 2, about 2.414.
Check
f(2) and f(3) both equal 5, and f(1 + √2) = 2 + 2√2 ≈ 4.83 dips below them.
Where this goes wrong
Picking 1 − √2 ≈ −0.414 because it is the smaller x, or because f is smaller there. It is a local maximum; the local minimum is the critical point on the right of the asymptote, where the branch turns from falling to rising.
The idea behind it
The curve sketching lesson, a figure to move, in a few steps
Leans on
The quadratic formula, from Functions, graphs and inverses
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